## Calculus (3rd Edition)

$0$
By dividing both denomainator and numerator by $n^{3/2},$ we have $$\lim_{n\to \infty}a_n=\lim_{n\to \infty} \frac{n}{\sqrt{n^3+1}}=\lim_{n\to \infty} \frac{1/n^{3/2}}{\sqrt{1+(1/n^3)}}=\frac{0}{1} =0,$$ and hence the sequence $a_n$ converges to $0$.